From a supply of identical capacitors rated 8 mF , 250 V, the minimum number of capacitors required to form a composite 16 mF , 1000 V is

Question

From a supply of identical capacitors rated 8 mF , 250 V, the minimum number of capacitors required to form a composite 16 mF , 1000 V is (A) 2 (B) 4 (C) 16 (D) 32

Tardigrade Admin · Accepted Answer

Let ' n ' such capacitors are in series and such ' m ' such branch are in parallel. ∴ 250 × n =1000 ∴ n =4....(i) Also (8/ n ) × m =16 ⇒ m =(16 × n /8)=8....(ii) ∴ No. of capacitor =8 × 4=32

Q. From a supply of identical capacitors rated $8 m F, 250 V$ , the minimum number of capacitors required to form a composite $16 m F, 1000 V$ is

2

4

16

32

Let ' $n$ ' such capacitors are in series and such ' $m$ ' such branch are in parallel.
$∴ 250 \times n = 1000$
$∴ n = 4$ ....(i)
Also $\frac{8}{n} \times m = 16$
$\Rightarrow m = \frac{16 \times n}{8} = 8$ ....(ii)
$∴$ No. of capacitor $= 8 \times 4 = 32$

Q. From a supply of identical capacitors rated 8mF,250V, the minimum number of capacitors required to form a composite 16mF,1000V is

2

4

16

32

Let ' n ' such capacitors are in series and such ' m ' such branch are in parallel. ∴250×n=1000 ∴n=4....(i) Also n8​×m=16 ⇒m=816×n​=8....(ii) ∴ No. of capacitor =8×4=32

Q. From a supply of identical capacitors rated $8 m F, 250 V$ , the minimum number of capacitors required to form a composite $16 m F, 1000 V$ is

Let ' $n$ ' such capacitors are in series and such ' $m$ ' such branch are in parallel.
$∴ 250 \times n = 1000$
$∴ n = 4$ ....(i)
Also $\frac{8}{n} \times m = 16$
$\Rightarrow m = \frac{16 \times n}{8} = 8$ ....(ii)
$∴$ No. of capacitor $= 8 \times 4 = 32$